Example of a pair which is not weakly annihilating

Question

Let $\mathcal F$ denotes the Fourier transform $\mathcal{F} :L^2(\mathbb R)\rightarrow L^2(\mathbb R)$ and $E, \Sigma$ be two measurable sets in $\mathbb R$.

The pair $(E,\Sigma)$ is called a weakly annihilating pair, if for any $f \in L^2(\mathbb{R})$, $support (f) \subseteq E$, $support(\mathcal F f)\subseteq \Sigma$, implies $f \equiv 0$.

The pair $(E,\Sigma)$ is called a strongly annihilating pair, if there exists a constant $C$ such that for any $f \in L^2(\mathbb{R})$,

$$\|f\|_2^2 \leq C \left(\int_{\mathbb R \setminus E} |f|^2 dx + \int_{\mathbb R \setminus \Sigma} |\mathcal F f|^2 d\xi \right).$$ The notion of annihilating pair arises in the study of uncertainty property in Fourier Analysis. For example Benedicks's Theorem says if $E$, $\Sigma$ are both sets of finite measures then they form a weakly annihilating pair, whereas Theorem of Amrein and Berthier says they form a strongly annihilating pair.

I am looking for examples of

1) A weakly annihilating pair which is $\underline{not}$ a strongly annihilating pair. ( Willie Wong has already answered this and I realise this was rather easy and I should have been able to figure it out myself, so my apologies.)

1') Sets $E$, $\Sigma$ both have infinite measure such that $(E,\Sigma)$ is a strongly annihilating pair.

2) Sets $E$, $\Sigma$ such that $E^c$ and $\Sigma^c$ have nonzero measure and $(E,\Sigma)$ is $\underline{not}$ a weakly annihilating pair.

I realise the standard reference for this topic is the book by Havin and Jöricke, which unfortunately our library does not have a copy of!! Is there any alternative reference someone can suggest ?

Thankyou.

Answer 1

I think the following may work for (1).

Let $E$ be the interval $[-1,1]$, and $\Sigma$ be the set $\mathbb{R}\setminus [-1,1]$. If $f$ has compact support, its Fourier transform can be extended analytically to $\mathbb{C}$, and so if $\mathcal{F}f$ vanishes on any interval ($\mathbb{R}\setminus \Sigma$), $f$ must be identically 0. So $E,\Sigma$ form a weakly annihilating pair.

Now consider an arbitrary odd Schwarz function $g$ (so that $g(x) = - g(-x)$) with $L^2$ norm 1. Write $\hat{g}$ for its fourier transform. Consider the family $g_\lambda(x) = \lambda^{1/2} g(\lambda x)$. It is clear that $g_\lambda$ is in $L^2$, and that as $\lambda\to\infty$ we have $\int_{E^c} g_\lambda^2 dx \to 0$.

One easily checks that $\hat{g_\lambda}(\xi) = \frac{1}{\lambda^{1/2}}\hat{g}(\frac\xi\lambda)$. We estimate $\int_{-1}^1 \hat{g}_\lambda^2 d\xi$ by $$2 \sup_{\Sigma^c} \hat{g}_\lambda^2 = \frac{2}{\lambda^{1/2}} \sup_{(-\lambda^{-1},\lambda^{-1})} \hat{g}$$ which using that $\hat{g}(0) = 0$ and its derivatives are uniformly bounded, gives that as $\lambda\to\infty$ the integral $\int_{\Sigma^c}\hat{g_\lambda}^2d\xi\to 0$ also.

Combining we have that there cannot be a constant $C$ for the "strongly annihilating condition" to hold.

Answer 2

1') As stated before, $\epsilon$-thin sets need not have finite measure, for instance, if $E=F = \bigcup\limits_{n \in \mathbb{Z}}[n-\frac{1}{200|n|},n+\frac{1}{200|n|}]$, then $(E,F)$ is strongly annihilating.

2) Take $E = F = \bigcup\limits_{n \in \mathbb{Z}} [n-\frac{1}{3},n+\frac{1}{3}]$. Their complements are clearly of infinite measure, yet notice that using Poission summation $\widehat{\sum\limits_{n} \delta_n} = \sum\limits_{n} \delta_n$ and convolving and multpilying by a compactly supported function $\varphi$ with small enough support, you get a counterexample to weak annihilation.